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\(25x^2-10xy+y^2=\left(5x\right)^2-2.5x.y+y^2=\left(5x-y\right)^2\)
\(\dfrac{4}{9}x^2+\dfrac{20}{3}xy+25y^2=\left(\dfrac{2}{3}x\right)^2+2.\dfrac{2}{3}x.5y+\left(5y\right)^2=\left(\dfrac{2}{3}x+5y\right)^2\)
a) 25x² - 16
= (5x)² - 4²
= (5x - 4)(5x + 4)
b) 16a² - 9b²
= (4a)² - (3b)²
= (4a - 3b)(4a + 3b)
c) 8x³ + 1
= (2x)³ + 1³
= (2x + 1)(4x² - 2x + 1)
d) 125x³ + 27y³
= (5x)³ + (3y)³
= (5x + 3y)(25x² - 15xy + 9y²)
e) 8x³ - 125
= (2x)³ - 5³
= (2x - 5)(4x² + 10x + 25)
g) 27x³ - y³
= (3x)³ - y³
= (3x - y)(9x² + 3xy + y²)
a) \(25x^2-16=\left(5x-4\right)\left(5x+4\right)\)
b) \(16a^2-9b^2=\left(4a-3b\right)\left(4a+3b\right)\)
c) \(8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
d) \(125x^3+27y^3=\left(5x+3y\right)\left(25x^2-15xy+9y^2\right)\)
e) \(8x^3-125=\left(2x-5\right)\left(4x^2-10x+25\right)\)
g) \(27x^3-y^3=\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
1. \(x^4-2x^2+1=\left(x^2-1\right)^2\)
2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
3. \(16x^2-8x+1=\left(4x-1\right)^2\)
4. \(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\)
5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)
6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)
7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)
Kết quả nè
1.(x^2-1)^2×(x^2+1)^2
2.(2x-5)^2÷4
3.(4x-1)^2
4.(x+y)^2×(x-y)
5.(1/4x-4/9y)×(1/4x+1/9y)
a)<=>
A,=(x+y)(x-y)=x^2-y^2
x=(-1/2)^5:(1/2)^4=-1/2
x^2=1/4
y=8^2/(-2)^5=-2
y^2=4
A=1/4-4=-15/4
a, \(\left(2x-1\right)^2-\left(3x-1\right)^2=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=-x\left(5x-2\right)\)
b, \(\left(x+1\right)^2-9=\left(x+1-3\right)\left(x+1+3\right)=\left(x-2\right)\left(x+4\right)\)
c, \(\left(4x-1\right)^2-9x^2=\left(4x-1-3x\right)\left(4x-1+3x\right)=\left(x-1\right)\left(7x-1\right)\)
d, \(x^2-9=\left(x-3\right)\left(x+3\right)\); e, \(x^2-25=\left(x-5\right)\left(x+5\right)\)
f, \(\left(x+2\right)^2-\left(3x-1\right)^2=\left(x+2-3x+1\right)\left(x+2+3x-1\right)=\left(-2x+3\right)\left(4x+1\right)\)
i, \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3-y^2\right)\left(x^3+y^2\right)\)
a) \(x(x+y)-y(x+y)\)
Ta có:
\(= (x-y)(x+y)\)
\(=x^2-y^2\)
Tính \(x,y\):
\(x=\left(-\dfrac12\right)^5:\left(\dfrac12\right)^4\)
\(=-\dfrac1{32}:\dfrac1{16}=-\dfrac12\)
\(y=8^2:(-2)^5\)
\(=64:(-32)=-2\)
Thay vào biểu thức:
\(A=\left(-\dfrac12\right)^2-(-2)^2\)
\(=\dfrac14-4\)
\(=-\dfrac{15}{4}\)
b) \((x-y)(x^2+xy+y^2)-(x+y)(x^2-y^2)\), biết \(x-y=0\).
Ta có:
\(x-y=0\Rightarrow x=y\).
\(B=0\cdot(x^2+xy+y^2)-(x+y)(x-y)(x+y)\)
\(=0-0\)
\(=0\).
c) \(x^3(x^2-y^2)+y^2(x^3-y^3)\)
Biết \(x=16^5:8^5:(-2)^4\) và \(|y|=1\).
Tính \(x\):
\(x=\dfrac{16^5}{8^5}:16\)
\(=2^5:16=32:16=2\).
Vì \(|y|=1\) nên \(y=1\) hoặc \(y=-1\).
Ta rút gọn biểu thức:
\(C=x^5-x^3y^2+x^3y^2-y^5\)
\(=x^5-y^5\).
- Nếu \(y=1\): \(C=2^5-1^5=31\).
- Nếu \(y=-1\): \(C=2^5-(-1)^5=33\).
Vậy \(C=31\) hoặc \(C=33\).
d) Câu d bạn ghi các giá trị:
- \(x=y=0\)
- \(x=y=1\)
- \(x=\dfrac12,\ y=-\dfrac32\)
- \(x=\sqrt4,\ y=\sqrt9\)
Nhưng chưa có biểu thức cần thay số. Bạn gửi lại biểu thức của câu d nhé.
6, \(x^2-1+2xy+y^2=\left(x+y\right)^2-1=\left(x+y-1\right)\left(x+y+1\right)\)
7, \(4x^2-12x+9-y^2=\left(2x-3\right)^2-y^2=\left(2x-3-y\right)\left(2x-3+y\right)\)
8, \(16x^2-4y^2+4y-1=16x^2-\left(2y-1\right)^2=\left(4x-2y+1\right)\left(4x+2y-1\right)\)
9, \(25-x^2-12x-36=25-\left(x+6\right)^2=\left(5-x-6\right)\left(5+x+5\right)=-\left(x+1\right)\left(x+10\right)\)
10, \(x^2-9-5\left(x+3\right)=\left(x-3\right)\left(x+3\right)-5\left(x+3\right)=\left(x+3\right)\left(x-8\right)\)
a) 25x² - 10xy + y²
= (5x)² - 2.5x.y + y²
= (5x - y)²
b) 4/9 x² + 20/3 xy + + 25y²
= (2/3 x)² + 2.2/3 x.5y + (5y)²
= (2/3 x + 5y)²
c) 9x² - 12x + 4
= (3x)² - 2.3x.2 + 2²
= (3x - 2)²
d) Sửa đề: 16u²v⁴ - 8uv² + 1
= (4uv²)² - 2.4uv².1 + 1²
= (4uv² - 1)²
chỉ cần giải mỗi c và d thui
giúp với sos
a) \(25x^2-10xy+y^2=\left(5x-y\right)^2\)
b) \(\dfrac{4}{9}x^2+\dfrac{20}{3}xy+25y^2=\left(\dfrac{2}{3}x+5y\right)^2\)
c) \(9x^2+12x+4=\left(3x+2\right)^2\)
d) \(16u^2v^4-8uv^2+1=\left(4uv^2-1\right)^2\)